More fun in 2015, Part 43

We have if $\omega$ is a primitive $2015^\text{th}$ root of unity then $\bar\omega$ is also a primitive $2015^\text{th}$ root of unity

We have $\frac{1}{1-\omega} = \frac{1}{\omega*\bar{\omega} - \omega} = \frac{-1}{\omega*(1-\bar{\omega})} = \frac{-\bar{\omega}}{1-\bar{\omega}}$

so every two item of the sum will "cancel" out and have the sum of 1.

We also know that the number of primitive roots of unity of $n$ is $\phi(n)$ , so the sum in our case is $\frac{\phi(2015)}{2} = \boxed{720}$

the primitive nth roots of unity are roots of the cytplomatic polynomial $\Phi_n(x)$ .read this .we know $\Phi_{2015}(x)=\prod (x-\omega)$ take log and differentiate both sides $\dfrac{\Phi_{2015}'(x)}{\Phi_{2015}(x)}=\sum \dfrac{1}{x-\omega}$ great. all we need to do is find $\dfrac{\Phi_{2015}'(1)}{\Phi_{2015}(1)}$ . we know that $x^n-1 = \prod_{d \mid n} \Phi_d(x).$ divide both sides by $\Phi_1(x)=x-1$ to find $x^{n-1}+x^{n-2}+...+x+1=\prod_{d \mid n,d>1} \Phi_d(x)$ take log and differentiate both sides $\dfrac{(n-1)x^{n-2}+...+2x+1}{x^{n-1}+x^{n-2}+...+x+1}=\sum_{d \mid n,d>1} \dfrac{\Phi_{d}'(x)}{\Phi_{d}(x)}$ put x=1 to simplify: $\dfrac{n-1}{2}=\sum_{d \mid n,d>1} \dfrac{\Phi_{d}'(1)}{\Phi_{d}(1)}=\sum_{d \mid n,d>1} \dfrac{\Phi_{d}'(1)}{\Phi_{d}(1)}u(\dfrac{n}{d})$ use möbius inversion: $\dfrac{\Phi_{n}'(1)}{\Phi_{n}(1)}=\sum_{d \mid n}\mu(\dfrac{n}{d})\dfrac{d-1}{2}=\frac{1}{2}\sum_{d \mid n}\left(\mu(\dfrac{n}{d})d\right)-\dfrac{1}{2}\sum_{d \mid n}\left(\mu(\dfrac{n}{d})\right)$ the first summation is equal to $\phi(n)$ , from one of otto sirs trig problem, and the 2nd is zero, as $\mu$ is the dirichlet inverse of u(unit function). (note, we are assuming n≠1). this all evaluates to $\dfrac{\Phi_{n}'(1)}{\Phi_{n}(1)}=\dfrac{\phi(n)}{2}$ put 2015 to get $\boxed{720}$ .